Logarithmic and Exponential Functions

Transcription

1 Asymptotes and Intercepts Logarithmic and exponential functions have asymptotes and intercepts. Consider the functions f(x) = log ax and f(x) = lnx. Both have an x-intercept at (1, 0) and a vertical asymptote at x = 0 (y-axis) but no horizontal asymptote. The function g(x) = a x (a 1) has a y-intercept at (0, 1) and a horizontal asymptote at y = 0 (x-axis), but has no vertical asymptote. Consider the following example: The graph of y = 2 x keeps getting closer and closer to the x-axis without touching it and there is a horizontal asymptote at y = 0. The graph of y = log2(x) keeps getting closer and closer to the y-axis without touching it and there is a vertical asymptote at x = 0. In general, the domain of y = logn P(x) is all x values such that P(x) > 0. Taking log of a negative number, or 0 for that matter, is not allowed. For example, if log(-2) or log(0) is evaluated on a calculator, an error message is 1

2 displayed. In addition, consider the following: The domain of y = log(x) is all x values larger than 0, or x > 0. The domain of y = log(x 3) is all x values such that x 3 > 0, or x > 3. The domain of y= a x is all real numbers because any real number can be substituted in for x. The domain of f(x)= 5e x and g(x) = 2.3 x is all real numbers. Logarithm and exponential graphs can be shifted, reflected, stretched, and shrunk just like other basic graphs. For example, if c > 0: h(x) = e x + c shifts the graph of f (x) = e x right c units. h(x) = log(x c) shifts the graph of f(x) = log(x) right c units. h(x) = e x +c shifts the graph of f(x) = e x up c units. h(x) = log(x + c) shifts the graph of f(x) = log(x) left c units. h(x) = -e x rotates the graph of f(x) = e x over the x-axis. h(x) = -log(x) rotates the graph of f(x) = log(x) over the x-axis. h(x) = e -x rotates the graph of f(x) = e x over the y-axis. h(x) = log(-x) rotates the graph of f(x) = log(x) over the y-axis. h(x) = ce x stretches the graph vertically if c > 1 and shrinks the graph vertically if 0 < c < 1. h(x) = clog(x) stretches the graph vertically if c > 1 and shrinks the graph vertically if 0 < c < 1. The graph of y = logb x has a vertical asymptote at x = 0 and an x-intercept at (1, 0). Therefore, g(x) = logb (x 1) will have an x-intercept at (2, 0) and a vertical asymptote at x = 1 because this is a shift right of 1 unit. 2

3 The graph of f(x) = 2 x has a y-intercept at (0, 1) and a horizontal asymptote at y = 0. Therefore, the graph of g(x) = 2 x 1 will have a y-intercept at (0, 0) and a horizontal asymptote at y = -1 because this is a shift down of 1 unit. Below is an example of an exponential function which calculates the amount of 3

4 money returned for deposit money into a bank account or certificate of deposit (CD) looks like this: A is the amount of returned. P is the principal amount deposited. r is the annual interest rate (expressed as a decimal). n is the compound period. t is the number of years. Suppose you deposit $5,000 at a rate of 5% compounded annually (n = 1). Then, the amount returned as a function of time is given by A = 5,000(1.05) 2. To determine the amount after 2 years, use A = 5,000(1.05) 2 = $5, The constant that has a value approximately equal to is e. It occurs many times in mathematics, especially in exponential functions. Constant environmental changes are examples where e will occur (the flow of blood, populations, and energy exchanges). In particular, it is common for a bank to compound interest continuously instead of a specific period of time. In this case, interest is changing constantly and the formula for calculating the amount is given by A = Pe rt. For example, suppose you deposit $5,000 at a rate of 5% compounded continuously, then the amount returned as a function of time is given by A= 5,000e 0.05t. To determine how much the amount is after 2 years, use A= 5,000e 0.05*2 = 5,000e 0.10 = $5,

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6 Steps 1. Get in logb M = N or ln M = N form. 2. Convert to exponential form: b N = M or e N = M. 3. Solve the resulting equation to get in x = number form. Suppose f(x) = 30 5lnx gives the number of children under 14 as a percentage of the entire population, f(x), x years after To find the year that the percentage was 20%, substitute 20 in for y = f(x) and solve for x. 20 = 30 5lnx Get in form first by subtracting the 30 from both sides and then dividing through by = -5lnx 2 = lnx e 2 = x x years Change to exponential form. Therefore, in the year , or approximately 1978, the percentage of the entire population that was under 14 was approximately 20%. 6